B(F(-),-) : B^op \times B \to Set

and A(GF(-),G(-))

and finally A(-,G(-))

and these are bifunctors, correct?

yes

and if you plug in the objects you get sets

i see

why are there three functors?

ah because they appeared in my answer ^^

I needed a functor as intermediate step

i see, so we need to show that each step each isomorphism was natural

yes

then the composition of the natural transformations is natural

(and bijective)

the proof of that is easy: you just draw the commutative squares next to each other to get a commutative rectangle

so can i try to understand the naturality first isomorphism

does it matter that the objects here are sets?

i think that is confusing me

it doesnt

it just happens to be that the target of our functors is the category of sets

A(-,-) : A^{op}\times A \to Set

so i have a diagram with four objects, which are morphisms

yes, morphisms in the category Set

the objects are a,b,c,d

hmm

not really

the objects are pairs

so you only have two objects

like (a,b) and (c,d)

i see

remember, we have a bifunctors

so i have (F(a), b) \rightarrow GF(a) \arrow Gb

and in the bottom row

(F(c) \arrow d) \rightarrow GFc \arrow Gd

and then i need vertical arrows, right?

yep

but its not obvious what those should be

now you may have to think a bit: what are the morphisms in the category A^{op}\times A

and what does A^{op} mean anyway?

pairs of morphisms, the first in A^op, the second in A

A^op just reverses the directions of the arrows

good

so you have a morphism (f,g) : (a,b) -> (c,d)

i.e. f : c -> a and g : b -> d

yes

so how do we get a morphism (F(a),b) -> (F(c),d) ?

that's (F(f), g)

in this case, because we are in Set, the morphism will just be a function

we know how to define functions, right? :D

yes

oh, it needs to compose with both coordinats

so (F, 1)

think more concretely

given a morphism h : F(a) -> b

we need to construct a morphism F(c) -> d

(now in the category B!)

we're given f : c -> a, g : b -> d and the functor F

i'm sorry, i don't see it

basically we need to construct this: F(c) -> F(a) -> b -> d

oh, its g \circ F(f)

well almost, u're missing the middle part

g \circ h \circ F(f)

yes

so what we've discussed is how the functor (F(-),-) works basically

that is, how it maps morphisms (a,b) -> (c,d) to morphisms in Set

that's how you get the vertical arrows

i think i see

i'll keep working on it

thank you so much for your help

np

and btw

if you know partially ordered sets

these are examples of categories

easy to work with

i'll look into those once i write down a solution to this problem

thanks

you can try to apply the concept of adjoint functors to those

functors are just monotonic maps

and adjoint functors are so called Galois connections

good luck

thanks

Hi @Topos and @Long. If I may ask, is this intended as a room to discuss topics from category theory in general?

I am asking partly because another user asked quite recently whether we could have room for category theory. Coincidentally, you have created this room, so maybe this room could serve that purpose...?

If this room is intended as a general category room it could be added into the List of chatrooms. (This might attract one or two users at least to peak in.)

Or we could wait a bit to see whether this room gets some activity and ony then add it to the list.

I will also add link to some experiments with writing commutative diagrams in chat.

Oh I definitely had no intention to start or even maintain a big channel. Topos needed some additional details which I couldn't fit into the answer, so we moved here. I'm not familiar with the chat system, but I suppose it could be beneficial to have a room for category theory. On the other hand I'm not currently learning category or am no expert, so I wouldn't task myself with kickstarting conversations here...

@Long Thanks for the response.

So I take that your answer is that this is not planned as the "main category theory" chatroom.

yes

But I guess that still if I (or somebody else) start such a room, we can let you know - maybe you could be interesting enough to peek in that room occasionally.

you seem to have some experience with chat rooms, what does it take to maintain one?

I'm just curious

There isn't really anything to maintain actually.

The only problem is to get enough users to get room working, so if you want the room to prosper, you probably need to mention it occasionally in main chat or elsewhere in order to attract some attention.

I have to say that most of the rooms devoted to some specific subject are not very active. But you can have a look at the rooms in this list and judge for yourself.

To be more specific, in order to get general topology chatroom started, I have repeatedly mentioned in the main chatroom. The room info says it has average around 330 messages per week, but this is probably calculated from relatively recent period. (There were rather long periods in this room with almost no activity.)

@MartinSleziak Since it doesn't look like you can delete a room, we may as well make it habitable for future uses

@Long Well, if you want to continue your discussion with Topos, I definitely do not want to interrupt.

So I can simply create a new room if that's the preferred way to go.

But if you already finished your discussion and you do not object, we can use this one.

Whichever way you prefer.

room topic changed to Category Theory: discuss abstract nonsense. draw and share diagrams with presheaf.com [category-theory] [learning]

Ok, I see form the room topic that we might try to make this a category theory room.

I guess you do not object if I add the room to the list I've mentioned before.

not in any way

Re: What does it take to maintain one? As I said, there's not really much maintenance involved. More important thing is that there is actually somebody who comes to the room and discusses stuff related to the subject.

Re: it doesn't look like you can delete a room. Actually, if a room is inactive (has no messages) for 14 days, than it gets frozen.

I have added this room to the List of chatrooms.